For problems 1–2, find the missing side length.
(Image: A right triangle with sides 9 in, 12 in and c.)
12 in.
15 in.
17 in.
21 in.
1. Using the Pythagorean theorem, we have:
c^2 = 9^2 + 12^2
c^2 = 81 + 144
c^2 = 225
c = √225
c = 15 in.
Therefore, the missing side length is 15 in.
Answer: B) 15 in.
2. This question is not solvable as there is no specific information given about which side length is missing.
2. (Image: A right triangle has a hypotenuse of 125 m and sides of 100 m and b.) (1 point)
25 m
50 m
75 m
100 m
Using the Pythagorean theorem, we have:
125^2 = 100^2 + b^2
15625 = 10000 + b^2
b^2 = 5625
b = √5625
b = 75 m
Therefore, the missing side length is 75 m.
Answer: C) 75 m.
2/sqrt5
To simplify 2/sqrt(5), we need to rationalize the denominator by multiplying both the numerator and denominator by sqrt(5):
2/sqrt(5) * sqrt(5)/sqrt(5) = 2sqrt(5)/5
Therefore, 2/sqrt(5) simplifies to 2sqrt(5)/5.
Answer: 2sqrt(5)/5.
-11sqrt112
To simplify -11√112, we first need to find the perfect square factor of 112. The prime factorization of 112 is:
112 = 2 × 2 × 2 × 2 × 7
We can group the 2's to get √(2 × 2 × 2 × 2 × 7) = 4√7.
So, -11√112 can be written as -11 × 4√7 = -44√7.
Therefore, -11√112 simplifies to -44√7.
Answer: -44√7.
17*sqrt17-9*sqrt17
To simplify 17√17 - 9√17, we can combine like terms:
17√17 - 9√17 = (17 - 9)√17
Simplifying the expression within the parentheses, we get:
(17 - 9) = 8
Therefore, 17√17 - 9√17 simplifies to 8√17.
Answer: 8√17.